Pumping Lemma

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Pumping Lemma Distinguishability

• Jim Hook Tim SheardPortland State University

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Importance of loops

• Consider this DFA. The input string 01011 gets
  accepted after an execution that goes through the
  state sequence s ? p ? q ? p ? q ? r. This path
  contains a loop (corresponding to the substring
  01) that starts and ends at p. There are two
  simple ways of modifying this path without
  changing its beginning and ending states

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• delete the loop from the path
• instead of going around the loop once, do it
  several times. As a consequence, we see that all
  strings of the form 0(10)i11 (where i ³ 0) are
  accepted.

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Long paths must contain a loop

• Suppose n is the number of states of a DFA. Then
  every path of length n or more visits at least
  n1 states, and therefore must visit some state
  twice. Thus, every path of length n or longer
  must contain a loop.

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The pumping lemma

• Suppose L is a regular language, w is a string
  in L, and u is a non-empty substring of w. Thus,
  wxuy, for some strings x, y . We say that u is a
  pump in w if all strings xuiy (that is, xy, xuy,
  xuuy, xuuuy, ) belong to L.
• Pumping Lemma. Let L be a regular language. Then
  there exists a number n, such that for all w Î L
  such that w ³ n, there exists a prefix of w
  whose length is less than n which contains a
  pump. Formally If w Î L and w ³ n then w
  xyz such that
• y ¹ e ( y is the pump)
• xy? n (xy is the prefix)
• xyiz Î L
• Definition. The number n associated to the
  regular language L as described in the Pumping
  Lemma is called the pumping constant of L.

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Proof

• w Î L, w ³ n, w xyz such that 1. y ¹ e 2.
  xy? n 3. xyiz Î L
• Let the DFA have m states. Let w³m. Consider
  the path from the start state s to the
  (accepting) state d(s,w). Just following the
  first m arcs, we make m1 total visits to states,
  so there must be a loop formed by some of these
  arcs.
• We can write wopqr, where p corresponds to that
  loop, andopq m (the prefix of size m). Thus
  let nop, xo, y p, and z qr.
• 1) Since every loop has at least one arc, we know
  p gt0, thus y ¹ e
• 2) xy? n because xy op and n op
• 3) xyiz Î L because If p is a loop, its starts at
  state si and
• d(si,p) si, and we know that d(si,qr)
  sfinal.. Thus
• d(sstart,x) si, Thus for each i d(si,yi)
  si, and were done.

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y q
x p
start
qi
final
z rs
r s
m steps
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Proving non-regularity

• To prove that a given language is not regular, we
  use the Pumping Lemma as follows.
• Assuming L is regular (we are arguing by
  contradiction!), let n be the pumping constant of
  L. Making no other assumptions about n (we don't
  know what it is exactly), we need to produce a
  string wÎL of length ³ n that does not contain a
  pump in its n-prefix. This w depends on n we
  need to give w for any value of n.
• There are many substrings of the n-prefix of our
  chosen w and we must demonstrate that none of
  them is a pump. Typically, we do this by writing
  wxuy, a decomposition of w into three substrings
  about which we can only assume that u ?e and
  xu ? n. Then we must show that for some
  concrete i (zero or greater) the string xuiy does
  not belong to L.

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Skill required

• Notice the game-like structure of the proof.
  Somebody gives us n. Then we give w of length ³
  n. Then our opponent gives us a non-empty
  substring u of the n-prefix of w (and with it the
  factorization w xuy of w). Finally, we choose i
  such that xuiyÏ L.Our first move often requires
  ingenuity We must find w so that we can
  successfully respond to whatever our opponent
  plays next.

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Example 1

• We show that L0k1k k0,1,2, is not
  regular. Assuming the Pumping Lemma constant of L
  is n, we take w0n1n. We need to show that there
  are no pumps in the n-prefix of w, which is 0n.
  If u is a pump contained in 0n then 0n xuz,
  and xuuz must also be in the language. But since
  u gt 0, if xuz n then xuuz m where m gt
  n. So we obtain a string 0m1n with mgtn, which is
  obviously not in L, so a contradiction is
  obtained, and are assumption that 0K1K is regular
  must be false.
• Note. The same choice of w and i works to show
  that the language
• Lw Î 0,1 w contains equal number of
  0s and 1s
• is not regular either.

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Example 2

• We show that L uu uÎa,b is not
  regular. Let n be the pumping constant. Then we
  choose wanbanb which clearly has length greater
  than n.
• The initial string an must contain the pump, u.
  So w xuybanb, and xuyb anb. But pumping u 0
  times it must be the case that xybanb is in L
  too. But since u is not e, we see that xyb ?anb,
  since it must have fewer as. Which leads to a
  contradiction. Thus our original assumption that
  L was regular must be false.
• Question. If in response to the given n we play
  wanan, the opponent has a chance to win. How?

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Example 3

• The language L w Î a,b,c the length of
  w is a perfect square is not regular. In
  response to n, we play any string w of length n2
  (which clearly has length greater than n). The
  opponent picks a pump u such that w xuy let
  ku and we have
• xuiy xuy (i-1) u n2 (i-1)k.
  
• If we can find i such that n2(i-1)k is not a
  perfect square, then we are led to a
  contradiction. A good choice is ikn21. In that
  case
• n2(i-1)k
• n2(kn21 1)k
• n2k2n2
• n2(k21), which is not a perfect square.

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Distinguishability
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Myhill Nerode

• The Myhill Nerode theorem is another
  characterization of the regular languages
• It uses a language to carve up the set of all
  strings into equivalence classes
• Intuitively these equivalence classes will
  correspond to states in a minimal DFA

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Definition

• x and y are distinguishable with respect to L if
  there is a z such that either xz is in L and yz
  is not in L or xz is not in L and yz is in L
• in other words
• x and y are indistinguishable wrt L if for all z,
  xz in L iff yz in L

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Example

• A a,b
• a and b are indistinguishable
• epsilon is distinguishable from all other strings
• all strings other than epsilon, a and b are
  indistinguishable
• In other words, there are three equivalence
  classes for A epsilon, a, aa.
• The number of equivalence classes induced by a
  language is called the index of the language (A
  is of index 3)

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Homework

• In homework you will show
• indistinguishable by L is an equivalence relation
• if L is recognized by a DFA with k states then L
  has index at most k
• If L has finite index k, then it is recognized by
  a DFA with k states
• L is regular iff it has finite index. The index
  is the size of the smallest DFA recognizing L

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Examples

• What is the index of anbn ?
• What are the equivalence classes of ab?
• What is the index of ab?